* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Calculating The Velocity of Gravity- and Capillarity
Survey
Document related concepts
Coandă effect wikipedia , lookup
Lift (force) wikipedia , lookup
Cnoidal wave wikipedia , lookup
Airy wave theory wikipedia , lookup
Fluid thread breakup wikipedia , lookup
Hydraulic machinery wikipedia , lookup
Flow measurement wikipedia , lookup
Compressible flow wikipedia , lookup
Computational fluid dynamics wikipedia , lookup
Flow conditioning wikipedia , lookup
Aerodynamics wikipedia , lookup
Navier–Stokes equations wikipedia , lookup
Derivation of the Navier–Stokes equations wikipedia , lookup
Reynolds number wikipedia , lookup
Transcript
Calculating The Velocity of Gravity- and Capillarity-Driven Flow A.J. Aranyosi March 16, 2010 1 Problem Description Imagine we have a microfluidic channel of length L, width w, and height h, with h << w << L. This device is initially empty, and a drop of liquid (e.g. blood) is placed at the opening of the device. The resulting flow is driven by a combination of forces due to gravity and capillarity. We would like to compute the corresponding velocity of this flow. 2 Pressure Due To Gravity Due to hydrostatic forces, the pressure at the bottom of the droplet of liquid is given by P = ρgH, where P is the pressure, ρ is the density of the liquid (roughly 1000 kg/m3 ), g is the gravitational constant (9.8 m/s), and H is the height of the liquid. For water and similar substances, P ≈ 104 H, with P in Pascals and H in meters. At atmospheric pressure, P ≈ 105 Pa, so H = 10 m; that’s a tall column of water! If we replace the water with mercury, at a density of 13,500 kg/m3 , the height drops to 0.75 m, or about 30 inches (the number you’re likely to hear in weather reports). For now, though, let’s assume that the volume of liquid that enters the channel is a negligible fraction of our initial droplet, so that H is constant. 3 Pressure Due To Capillary Action Capillarity is a force that results from the interaction of cohesion of molecules of a liquid to each other and adhesion of these molecules to the surrounding material. For example, water adheres tightly to glass, and so “climbs up” the walls of a glass tube. As the water on the edges of the tube moves up, the surface tension resulting from molecular cohesion of water pulls the center up as well, resulting in a positive capillarity force that moves the water further along the tube. Mercury, in contrast, is highly cohesive but does not adhere well to glass, so the molecules near the glass tend to be attracted toward other 1 mercury, resulting in a concave downward shape. This shape (which can be seen in a mercury thermometer), combined with the surface tension of mercury, results in a negative capillarity force that tends to pull the mercury downwards. In general, the pressure due to capillarity in a vertical cylindrical tube of radius r can be described by 2γ cos(θ) , (1) P = r where γ is the surface tension between the liquid and air (in J/m2 ; γ = 0.0728J/m2 for water at 20◦ C), θ is the contact angle (in radians, measured from the downward vertical, so water has θ = 20◦ or 0.35 rad while mercury has θ = 140◦ for contact with glass) between the liquid and wall, and r is the radius of the tube. For the case of a rectangular channel, the capillary force is dominated by the smallest dimension, so we can replace r with h in our calculations. 4 Resistance To Flow In lecture we derived an equation describing one-dimensional static Poiseuille flow, ∂2u ∂P = µ 2, (2) ∂x ∂y where P is the applied pressure, x is the dimension along the length of the channel, µ is the fluid viscosity, u is the fluid particle velocity (as distinguished from the volume velocity), and y is the dimension across the channel. With no-slip boundary conditions (i.e., ∂u ∂y is finite so u(0) = u(h) = 0), it is easy to show that u(y) = cy(h − y), where c is a constant. The peak velocity umax = Rh u(h/2) = ch2 /4, so c = 4umax /h2 . Similarly, u = h1 0 u(y)dy = ch2 /6 = 23 umax , so u(y) = 6u h2 y(h − y). Substituting this back into equation 2 gives ∂P = −12µu/h2 . ∂x 5 (3) Putting It All Together Now we’ll combine all of these derivations to determine how flow will be established in our microchannel. We will derive an equation for u as a function of l, the distance over which fluid has traveled, and use this equation to derive l(t). Here’s where we get to play fast and loose with physics. We’re going to assume that the flow in the microchannel is Poiseuille, despite the fact that capillarity will dominate the shape of the flow near the front. This assumption will probably make our answer wrong at a detailed level, but we’re not likely to be off by more than a factor of 2–3, and we should get the scaling factors correct. If we assume that our applied gravitational and capillary pressure act over the distance l, then we can substitute these into equation 3 to get (ρgH + 2γ cos θ )/l(t) = −12µu/h2 , h 2 (4) which gives u= h2 2γ cos θ (ρgH + ). 8µl(t) h (5) In other words, our fluid velocity will be dominated by the larger of the gravitational and capillary forces, will scale as the square of the smallest dimension h of the channel, and will be inversely proportional to the distance already traveled — i.e., the flow will slow down as it moves further along. This result makes intuitive sense: the driving pressures are essentially constant, but they must act on an ever-increasing volume of fluid. ∂l . With this Finally, we can use equation 5 to solve for l(t), since u = ∂t substitution, ∂l h2 2γ cos θ 1 = (ρgH + ) , (6) ∂t 8µ h l(t) or ∂l (7) l(t) = C, ∂t where C = h2 8µ (ρgH + 2γ cos θ ) h is constant. A simple solution to this nonlinear √ differential equation is l(t) = At1/2 , so A = 2C, which gives s th2 2γ cos θ l(t) = (ρgH + ). (8) 4µ h So the length traveled by the fluid will scale with the square root of time and applied pressure, inversely with the square root of viscosity, and linearly with the channel height (for gravity-driven flow) or the square root of channel height (for capillarity-driven flow). Consequences Several facts are evident from these equations. First, the force of gravity will exceed that of capillarity under two conditions. The first condition is when the channel has already been filled, either due to the capillary force or applied pressure; in this case, only the pressures due to gravity and applied forces will cos θ θ ; i.e., H > 2γρgh . For drive fluid flow. The second case is when ρgH > 2γ cos h water interfacing with a hydrophilic surface like glass or recently plasma-treated −5 . If h = 50 µm, PDMS, γ = 0.0728J/m2 and θ = 0.35 rad, so H > 1.4×10 h the gravitational force is less than capillarity for H < 28 cm, and becomes completely negligible for H < 28 mm. Put another way, the maximum H for a microfluidic device on a glass slide is 75 mm; for this H, the gravitational force is less than capillarity for a device height h < 187 µm, and becomes completely negligible for h < 19 µm. Thus even for the longest device we can make, the gravitational pressure for a 50 µm high channel will only be 1/3 as big as that of capillarity — so we can ignore gravity as a driving force in filling the channel. 3 The second consequence has to do with the length of the channel. If we want the device to be completely filled within one minute, for example, equation 8 gives r 60 ∗ (5 × 10−5 )2 2 × 0.0728 × cos 0.35 l(t) = = 0.32 m. (9) 4 × 10−3 5 × 10−5 So for a channel length of 0.2 m, the device will fill in about 48 seconds; for a channel length of 0.4 m, it will fill in about 67 seconds. In other words, a reasonable time to fill a reasonable-length channel is about a minute. We can vary this time somewhat by changing the length and/or the thickness of the channel, but a minute is a good rough guess. 4